Two bullets of equal mass were fired toward a target at the same speed but in opposite directions. The target was formed of two different pieces of metal stuck together. The first was nine centimeters thick, and the second was 12 centimeters thick. When the bullets hit the target, the first one passed through the first layer and embedded four centimeters into the second layer before it stopped, whereas the other bullet passed through the second layer and embedded five centimeters into the first layer before it stopped. Using the work–energy principle, calculate the ratio of the resistance of the first metallic layer to that of the second.
The diagram below represents a visualization of the problem, with the yellow box being the first layer with a thickness of nine centimeters and the pink box representing the second layer with a thickness of 12 centimeters. The top blue dotted line represents the first bullet, which ended up being embedded four centimeters into the second layer. And the bottom blue dotted line represents the second bullet, which ended up being embedded five centimeters into the first layer.
The problem tells us to use the work–energy principle. In equation form, that is that the net work done on an object is equal to the change in kinetic energy of the object. We can apply this equation to both bullets. Recall that the net work done on an object is equal to the force times the displacement. The net work done on bullet one is 𝐹 one times nine plus 𝐹 two times four. 𝐹 one represents the resistance of layer one and nine is the thickness as the bullet goes completely through layer one. And 𝐹 two represents the resistive force of layer two and four for the distance as the bullet is embedded four centimeters into layer two. We can leave the change in kinetic energy as ΔKE, as both bullets have the same mass, the same initial speed, and eventually came to rest.
The expanded equation for the second bullet is 𝐹 one times five plus 𝐹 two times 12 equals ΔKE. 𝐹 one and 𝐹 two once again represent the resistive force of the respective layers, layer one and layer two. The five is the distance that the second bullet gets embedded into the first layer, five centimeters. And the 12 comes from the fact that the bullet goes all the way through the second layer, which is 12 centimeters thick. We wanna find the ratio of 𝐹 one to 𝐹 two or the resistance of the first metallic layer to that of the second. To do this, we can subtract our equations. 𝐹 one times nine minus 𝐹 one times five is 𝐹 one times four. And 𝐹 two times four minus 𝐹 two times 12 is negative 𝐹 two times eight. When we subtract our two change in kinetic energies, we get zero, as the bullets are the same mass, same initial speed, and come to rest.
Our next step is to add 𝐹 two times eight to both sides of the equation. To rearrange our equation so that we have a ratio, we can divide both sides by 𝐹 two and both sides by four. The ratio of 𝐹 one to 𝐹 two is two to one, more commonly written in this form. The ratio of the resistance of the first metallic layer to that of the second is two to one.